/*
 * @lc app=leetcode.cn id=21 lang=java
 *
 * [21] 合并两个有序链表
 *
 * https://leetcode-cn.com/problems/merge-two-sorted-lists/description/
 *
 * algorithms
 * Easy (66.68%)
 * Likes:    2129
 * Dislikes: 0
 * Total Accepted:    816K
 * Total Submissions: 1.2M
 * Testcase Example:  '[1,2,4]\n[1,3,4]'
 *
 * 将两个升序链表合并为一个新的 升序 链表并返回。新链表是通过拼接给定的两个链表的所有节点组成的。 
 * 
 * 
 * 
 * 示例 1：
 * 
 * 
 * 输入：l1 = [1,2,4], l2 = [1,3,4]
 * 输出：[1,1,2,3,4,4]
 * 
 * 
 * 示例 2：
 * 
 * 
 * 输入：l1 = [], l2 = []
 * 输出：[]
 * 
 * 
 * 示例 3：
 * 
 * 
 * 输入：l1 = [], l2 = [0]
 * 输出：[0]
 * 
 * 
 * 
 * 
 * 提示：
 * 
 * 
 * 两个链表的节点数目范围是 [0, 50]
 * -100 
 * l1 和 l2 均按 非递减顺序 排列
 * 
 * 
 */

// @lc code=start
/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {

    // 版本 1.0
    public ListNode mergeTwoLists(ListNode list1, ListNode list2) {
        if (list1 == null || list2 == null) {
            return list1 == null ? list2 : list1;
        }
        // 新表头先指向list1和list2头节点小的那一个
        ListNode newListHead = list1.val < list2.val ? list1 : list2;
        if (newListHead == list1) {
            list1 = list1.next;
        } else {
            list2 = list2.next;
        }
        ListNode temp = newListHead;
        while (list1 != null || list2 != null) {
            if (list1 == null) {
                temp.next = list2;
                break;
            } else if (list2 == null) {
                temp.next = list1;
                break;
            } else if (list1.val < list2.val) {
                temp.next = list1;
                list1 = list1.next;
                temp = temp.next;
            } else {
                temp.next = list2;
                list2 = list2.next;
                temp = temp.next;
            }
        }
        return newListHead;
    }

    // 版本 1.5
    public ListNode mergeTwoLists(ListNode list1, ListNode list2) {
        ListNode newListHead = new ListNode();
        ListNode temp = newListHead;
        while (list1 != null || list2 != null) {
            if (list1 == null) {
                temp.next = list2;
                break;
            } else if (list2 == null) {
                temp.next = list1;
                break;
            } else if (list1.val < list2.val) {
                temp.next = list1;
                list1 = list1.next;
                temp = temp.next;
            } else {
                temp.next = list2;
                list2 = list2.next;
                temp = temp.next;
            }
        }
        return newListHead.next;
    }

    // 递归解法... 待补充...
}
// @lc code=end

